Problem: $\vec v = (8,-16)$ $-\dfrac14\vec v= ($
Solution: In general, the scalar multiple of $k$ times $\vec u$ is this: $k\vec u = k(u_x, u_y) = (ku_x, ku_y)$. So, here's how we find $-\dfrac14 \vec{v}$ : $\begin{aligned} {-\dfrac14}\vec v = {-\dfrac14} \cdot (8,-16) &= \left({-\dfrac14} \cdot 8, {-\dfrac14} \cdot (-16)\right) \\\\ &= (-2,4) \end{aligned}$ The answer is $ (-2,4) $.